Absolute Min and Max of f(x,y) = y sqrt x - y^2 - x + 3y

[chucknorrisfish]

Find the absolute maximum and minimum of the function f(x,y) = y sqrt x - y^2 - x + 3y on the domain 0 <= x <= 9, 0 <= y <= 7. Find the points that they are attained.

Here's what ive started:

f(x,y) = yx^.5 - y^2 - x + 3y
fx = (.5x^-.5 )(y) - 1
fy = (x^.5) - 2y + 3
fxx = (-.25x^-1.5)(y)
fyy = -2
fxy = .5x^-.5

So then i think i set fx and fy to 0.
fx = (1/2x^.5 )(y) - 1 = 0
fy = (x^.5) - 2y + 3 = 0

but thats where i get stuck, i guess im just not sure how to solve for x and y...

 

[Deleted member 4993]

Re: Calculus-Absolute Min and Max

Find the absolute maximum and minimum of the function f(x,y) = y sqrt x - y^2 - x + 3y on the domain 0 <= x <= 9, 0 <= y <= 7. Find the points that they are attained.

Here's what ive started:

f(x,y) = yx^.5 - y^2 - x + 3y
fx = (.5x^-.5 )(y) - 1
fy = (x^.5) - 2y + 3
fxx = (-.25x^-1.5)(y)
fyy = -2
fxy = .5x^-.5

So then i think i set fx and fy to 0.
fx = (1/2x^.5 )(y) - 1 = 0...................................(1)
fy = (x^.5) - 2y + 3 = 0......................................(2)
but thats where i get stuck, i guess im just not sure how to solve for x and y...

Assuming you done the problem correctly upto this point,
from (1)

y = 2x^(0.5)

y^2 = 4x

x = y^2/4......................................(3)

Substituting in (2)

y/2 - 2y + 3 = 0

Now you can solve for 'y' and use (3) to solve for 'x'. Don't forget to check your answer.